[bzoj5337][loj2576][TJOI2018]str【字符串哈希】【后缀自动机】【dp】

【题目链接】
　　https://www.lydsy.com/JudgeOnline/problem.php?id=5337
　　https://loj.ac/problem/2576
【题解】
　　随便怎么做都可以。
　　比较简单的做法是：记$f[i][j]$表示考虑了前$i$个串，当前的末尾匹配到了第$j$位。那么可以用字符串哈希判断一段是否能匹配做到$O(1)$转移。
　　时间复杂度$O(L*\sum A)$
　　但我做的时候脑回路比较清奇，把状态记到了后缀自动机上。。转移变成在自动机上游走。
　　奇怪的复杂度$O(L*\sum 子串长度)$
【代码】

# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    N       30010
# define    K       110
using namespace std;
int read(){
    int tmp = 0, fh = 1; char ch = getchar();
    while (ch < '0' || ch > '9') {if (ch == '-') fh = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9'){tmp = tmp * 10 + ch - '0'; ch = getchar(); }
    return tmp * fh;
}
struct Trie{
    int son[26], step, pre;
}T[N];
const int  P = 1e9 + 7;
int f[K][N], n, l, num[K], place, use[N], sum[N], tag[N], las;
char s[N], tmp[N];
int extend(char x, int l, int las){
    x -= 'A';
    int now = ++place;
    T[now].step = l;
    while (T[las].son[x] == 0){
        T[las].son[x] = now;
        if (las == 0){
            T[now].pre = 0;
            return now;
        }
        else las = T[las].pre;
    }
    int q = T[las].son[x], p = las;
    if (T[p].step + 1 == T[q].step)
        T[now].pre = q;
        else {
            int np = ++place;
            T[np].step = T[p].step + 1;
            T[np].pre = T[q].pre;
            for (int i = 0; i < 26; i++)
                T[np].son[i] = T[q].son[i];
            T[q].pre = np, T[now].pre = np;
            for (; T[p].son[x] == q; p = T[p].pre)
                T[p].son[x] = np;
        }
    return now;
}
void solve(int x){
    if (use[x]) return;
    use[x] = true;
    if (tag[x]) sum[x] = 1;
    for (int i = 0; i < 26; i++){
        if (T[x].son[i] == 0) continue;
        solve(T[x].son[i]);
        sum[x] = (sum[x] + sum[T[x].son[i]]) % P;
    }
}
int main(){
//  freopen("A.in", "r", stdin);
//  freopen(".out", "w", stdout);
    n = read();
    scanf("\n%s", s + 1);
    l = strlen(s + 1), las = 0;
    for (int i = 1; i <= l; i++)
        las = extend(s[i], i, las);
    while (las != 0) tag[las] = true, las = T[las].pre;
    f[0][0] = 1;
    for (int i = 1; i <= n; i++){
        num[i] = read();
        for (int j = 1; j <= num[i]; j++){
            scanf("\n%s", tmp + 1);
            int len = strlen(tmp + 1);
            for (int k = 1; k <= len; k++) tmp[k] = tmp[k] - 'A';
            for (int k = 0; k <= place; k++){
                if (f[i - 1][k] == 0) continue;
                int now = k, flag = true;
                for (int t = 1; t <= len; t++){
                    if (T[now].son[tmp[t]] == 0){
                        flag = false;
                        break;
                    }
                    now = T[now].son[tmp[t]];
                }
                if (flag == true){
                    f[i][now] = f[i][now] + f[i - 1][k];
                    f[i][now] = (f[i][now] >= P) ? (f[i][now] - P) : f[i][now];
                }
            }
        }
    }
    int ans = 0;
    solve(0);
    for (int i = 0; i <= place; i++)
        ans = (ans + 1ll * f[n][i] * sum[i]) % P;
    printf("%d\n", ans);
    return 0;
}